Question

Sketch the curve with the given vector equation. Indicate with an arrow the direction in which $t$ increases. $\mathbf{r}(t)=t^2 \mathbf{i}+t^4 \mathbf{j}+t^6 \mathbf{k}$

   Sketch the curve with the given vector equation. Indicate with an arrow the direction in which $t$ increases.
$\mathbf{r}(t)=t^2 \mathbf{i}+t^4 \mathbf{j}+t^6 \mathbf{k}$
Single Variable Calculus: Early Transcendentals
Single Variable Calculus: Early Transcendentals
James Stewart,… 9th Edition
Chapter 13, Problem 15 ↓
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Sketch the curve with the given vector equation. Indicate with an arrow the direction in which $t$ increases. $\mathbf{r}(t)=t^2 \mathbf{i}+t^4 \mathbf{j}+t^6 \mathbf{k}$
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Transcript

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00:01 So we have r of t equals the vector t squared, t to the fourth, t to the sixth.
00:09 All the coordinates are even powers.
00:11 So x equals t squared is greater than are equal to zero.
00:14 So is y equals two to the fourth and z equals t to the six.
00:17 So the whole curve lives in the first octant, which is the positive corner.
00:23 So let's eliminate the parameter.
00:26 So because x equals t squared, y is equal to t to the fourth would be t squared squared squared, which would be x squared, and then z equals t to the six, which is t to the second cubed, which would be then x cubed.
00:43 So the curve lies on the space, the space curve defined by y equals x squared, z equals x cubed where x is greater than equal to zero.
00:53 What this ends up looking like, and i'll do my best, but you would be better suited using a 3d calculator...
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